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In the book "Real and Complex Analysis" by Rudin, he often uses the condition that a space is locally compact Hausdorff in order to present results in a general manner. The thing is, I'm not very used to such condition. Most books of analysis/measure-theory that I've read present results in terms of metric spaces/separable/complete.

Thus, I was wondering if there is a precise relation between such notions. For example, does locally compact Hausdorff implies completeness or separability? Is the opposite implication true?

Take for example the following theorem by Rudin:

If $X$ is locally compact Hausdorff, and $\mu$ a measure on the borelians of $X$. Then for $1\leq p < \infty$, $C_c(X)$ is dense in $L^p(\mu)$.

Now, I was wondering if this theorem could be somehow stated, but something like, if $X$ is Polish, then this is true. Hence, what I'm really interested in is to know if there is a way to somehow relate this type of spaces. If the implications are not true, is there an extra condition that tie them together?

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    $\begingroup$ Perhaps you are looking for Urysohn's metrization theorem $\endgroup$
    – Bumblebee
    Nov 21, 2021 at 14:46
  • $\begingroup$ Thanks! I'll take a look. $\endgroup$ Nov 21, 2021 at 14:49
  • $\begingroup$ In your example, I guess $C_c(X)$ is the space of continuous functions with compact support. That is not a very interesting thing unless $X$ is locally compact. $\endgroup$
    – GEdgar
    Nov 21, 2021 at 15:01
  • $\begingroup$ What @Bumblebee posted is the sort of thing I was looking for! Thanks again! "Urysohn's Theorem can be restated as: A topological space is separable and metrizable if and only if it is regular, Hausdorff and second-countable. The Nagata–Smirnov metrization theorem extends this to the non-separable case. It states that a topological space is metrizable if and only if it is regular, Hausdorff and has a σ-locally finite base." $\endgroup$ Nov 21, 2021 at 15:06
  • $\begingroup$ Sorry if the question was not very precise. I guess it was one of those "you'll know when you see it". $\endgroup$ Nov 21, 2021 at 15:06

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Both implications fail.

Product space $[0,1]^A$ with $A$ uncountable is compact Hausdorff but not separable and not metrizable.

Hilbert space $l_2$ is complete separable metric, but not locally compact.

Of course, many common spaces have both properties. Indeed, an open subset of $\mathbb R^n$ is completely metrizable separable locally compact Hausdorff.

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  • $\begingroup$ Thanks for the answer. But is there a way to connect both properties? I mean, is there another condition they tie them together? Cause Rudin proves stuff assuming locally compact Hausdorff, and I see the same theorem proved somewhere else but without such assumption. $\endgroup$ Nov 21, 2021 at 14:49
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    $\begingroup$ Why not include such an example in the question? $\endgroup$
    – GEdgar
    Nov 21, 2021 at 14:51
  • $\begingroup$ Ok, done. Hopefully this makes things clearer on what I'd want to know $\endgroup$ Nov 21, 2021 at 14:56
  • $\begingroup$ For completeness (subset of $\mathbb{R}^n$), it would have to be closed. :-) $\endgroup$ Nov 21, 2021 at 17:39
  • $\begingroup$ @AndréCaldas ... good point. Or say "completely metrizable" then open sets are OK. Fixed $\endgroup$
    – GEdgar
    Nov 21, 2021 at 19:13
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Let $I=[0,1]$ with the standard topology. Let $k$ be an infinite cardinal. By the Tychonoff Theorem ( a product of compact spaces is compact), the product-space $I^k$ is compact.

It is easy to show that a product of $T_2$ spaces is $T_2$ and it is easy to show that a compact $T_2$ space is $T_4$. So the "Tychonoff plank" $I^k$ is a compact normal space. It is also easy to show that any subspace of a normal space is a $T_{3\frac 1 2}$ space.

Theorem: If $S$ is a $T_{3\frac 1 2}$ space and if $S$ has a base (basis) $B$ with cardinal $|B|\le k$ then $S$ is homeomorphic to a subspace of $I^k.$

So the class of compact Hausdorff spaces and their subspaces is, in this sense, much bigger than the class of metrizable spaces.

In particular a separable metrizable space has a countable base so it is homeomorphic to a subspace of $I^{\aleph_0}.$

It is hard to define a useful countably-additive measure on the Borel sets of a space that is not locally compact. For example in an infinite-dimensional normed linear space (e.g. Hilbert space $\ell_2$ ) there exists $ r>0$ such that an open ball of radius $1$ contains an infinite pairwise-disjoint family of open balls, each of radius $r$.

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  • $\begingroup$ The space $I^{\aleph_0}$ is called the Hilbert cube. It is metrizable. Its subspace $\{0,1\}^{\aleph_0}$ is homeomorphic to the Cantor set. $\endgroup$ Nov 21, 2021 at 17:43

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